Question

我有我的怀疑:

$ziua = "SELECT DISTINCT DAYOFMONTH(ziua) FROM rapoarte"; $ziuaResult = mysql_query($ziua);

With the results i get, i want to add them in a drop down box, like this: ` echo"Selectati Ziua:

            <td><select name= ziua >
            <option value= --- >---</option>";
        while($ziuaRow = mysql_fetch_array($ziuaResult)) {
         $ziua1 = $ziuaRow[ ziua ];
         echo "<option value= $ziua1 >$ziua1</option>";
        }

a. 重复”

`

问题在于,我的落户箱是空的。有2,3个备选办法(视特定结果而定),但没有案文。

I have the same problem with this selection: SELECT DISTINCT HOUR(ora) FROM rapoarte

如何解决这一问题?

thanks, Sebastian

http://www.ohchr.org。

sorry, i added the wrong code.

Answer 1

You are only selecting HOUR(ora) without the column ziua. The following should select the ziua column, unique by HOUR(ora).

SELECT DISTINCT HOUR(ora) AS something,ziua FROM rapoarte GROUP by something

如果您改名为error_ reporting(E_ALL);,你将发现关于未界定指数的错误,ziua。

Answer 2

您的论点是否应当传到我的sql_fetch_array() ziuaResult而不是$trunchiResult?

Answer 3

这一工作应当:

echo "<option value= ".$trunchi1." >".$trunchi1."</option>";

友情链接